PS 7 Chem 101 Spring 2025 2
Problem Set Overview
General Information
Course: Chem 101
Problem Set: 7
Semester: Spring 2025
Name line for student: ______________________________________
Electron Configuration Questions
Write the most stable electron configuration for:
(a) Sulfur (S, Z=16)
Configuration: [Ne] 3s² 3p⁴
(b) Technetium (Tc, Z=43)
Configuration: [Kr] 5s² 4d⁵
(c) Silver (Ag, Z=47)
Configuration: [Kr] 5s¹ 4d¹⁰
Electron Configuration for Ions
Electron configuration for Rh³⁺
Configuration: [Kr] 4d⁵
Note: Transition metals lose outermost s-electrons first.
Ionization Energy Change
Change in electron configuration associated with the first ionization energy of Aluminum (Al)
Option (a): [Ne] 3s²3p¹ → [Ne] 3s¹3p¹ + e⁻
This represents the removal of one electron.
Partial Orbital Diagrams
Identify neutral atom and +2 ion using the orbital diagram:
(a) represents a neutral atom, (b) is a +2 ion.
Periodic Trends
Order elements by increasing atomic radius (smallest to largest):
Elements: Cs, Rb, Sr
Order: Sr < Rb < Cs
Order elements by increasing ionization energy (least to greatest):
Elements: Al, B, C
Order: Al < B < C
Order elements by increasing electron affinity (least to greatest):
Elements: I, Rb, Cl
Order: Rb < I < Cl
Circle the atom or ion with larger radius in each pair:
(a) Ne vs Kr: Kr
(b) P vs P³⁻: P³⁻
(c) Al vs Al³⁺: Al
(d) Na⁺ vs F⁻: F⁻
(e) Cu⁺ vs Cu²⁺: Cu⁺
(f) O²⁻ vs S²⁻: S²⁻
Unpaired Electrons
Count unpaired electrons and identify paramagnetic atoms:
Molybdenum (Mo): ____
Nitrogen (N): ___
Vanadium (V): ____
Oxygen anion (O¹⁻): ___
Vanadium cation (V³⁺): ____
Additional Electron Configuration Questions
Change in electron configuration associated with the electron affinity of Aluminum (Al):
Option (c): [Ne] 3s²3p¹ + e⁻ → [Ne] 3s²
Count the number of valence electrons for each element:
Barium: 2
Phosphorus: 5
Lithium: 1
Germanium: 4
Iodine: 7
Sulfur: 6
Al²⁺: 3
Cl⁻: 8
Identify the main group element in Period 4 based on its valence shell orbital diagram:
Symbol:
Existence of an atom with a specific orbital diagram:
Can it exist?
If yes, symbol for the element:
Symbol and charge for the following cases:
(a) Element that lost one electron:
(b) Element that gained two electrons:
Cation formation by losing one electron:
Circle the electron to be lost:
Cation symbol and charge:
Identify the elements for the given electron configurations:
(a) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d⁵:
(b) [Xe]6s²4f¹⁵5d¹⁰6p⁵:
(c) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s¹4f¹⁵5d¹⁰:
(d) [Rn]7s²5f⁷6d¹:
Quantum numbers for 17 electrons of Chlorine (Cl):
(Completed example) 1, 0, 0, ½
Remaining configurations:
[Ne] 3s² 3p³
Quantum numbers for remaining electrons as listed.